Romanian Master of Physics 2017
Experimental problem (20 points)
Polycrystals and monomolecular layers
Experimental problem suggests you that using only one calibrated measuring instrument – a stopwatch –
and a few other simple objects and devices, to determine various characteristics of polycrystalline gold,
of the monolayers of stearic acid, of the carbon atom and to estimate the value of a fundamental physical
constant.
Fill corresponding boxes in attached answers sheets with results of all your taken measurements and all
your answers to work tasks.
Experimental apparatus
To solve experimental problem, you may use:
1. A stopwatch
2. A sheet of paper marked with a grid (marked lines are equidistant but the distance between them
is unknown).
3. A sheet of clear plastic that is marked with the same kind of grid as the paper.
4. A plummet.
5. A plate of expanded polystyrene that can be fixed on the work table.
6. A pin.
7. An electron diffraction image obtained for a polycrystalline layer of gold.
8. Two test tubes, one of which contains a solution. Do not drink the liquid in the tube!
9. Pipette
10. A sheet of transparent plastic.
11. Several sheets of white paper.
12. Marker.
13. Wipes.
14. A vessel for collection of waste liquids
It is forbidden the use of tools or measurement tools that are not included in the above list.
Do not drink the liquid in the tube!
It is not required an error calculation for none of working tasks.
1. Task A
Task A suggests you to use some of the items available, to determine the distance in millimeters
between the lines on the paper sheet (2) and the plastic sheet (3). This distance is called hereinafter
arbitrary unit of length, a.u.. The gravitational constant value is g  9,8 m  s 2 .
1.1. Requirements for task A
A.1. Designs an experiment to determine, in millimeters, the distance between the lines on the paper
sheet (2) and the plastic sheet (3) using some of the available items. Fill the appropriate box in the
answer sheet with a brief description of the experimental method that you propose.
A.2. Make measurements as in proposed method, and record measured data in the appropriate tables in
the answer sheet.
A.3. Use a suitable processing of data obtained from measurements and determine (in mm.) the value of
arbitrary unit of length - the distance between lines on the sheet of paper and plastic sheet. In the
appropriate box in the answer sheet, specify the method that you have chosen for processing data,
describe the calculations for determining the value of a.u. in millimeters and write the result in the format
#, # mm.
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In the following you will use grid of paper sheet and plastic sheet for measuring the distance in the next
working tasks.
2. Task B
In this task it is proposed to analyze diffraction image which you were made available and to respond to
the requirements of this part of the experimental problem.
We recommend you to carefully read the paragraph 2.1.
2.1. About polycrystals's electron diffraction image.
For crystalline substances, which are stable solids, arrangement of the constituents (atoms or
molecules) is characterized by an orderly position in space. Regular structure of a crystalline solid is
characterized by the existence of a periodicity that manifests itself by reproducing its structure after
equal distances along any straight line passing through atoms or molecules belonging crystal. A

parallelepiped built on three fundamental vectors ai is called a unit cell of the crystal lattice. By
translating with any combination of fundamental vectors of the unit cell can reconstitute the entire crystal
lattice. Between the crystal lattices, some have a cube as elementary cell. Side of the cube, a , is called
network constant.
(A)
(B)
(C)
In the above figure they are shown three types of elementary cubic cells. Cell of Simple Cubic network,
S. C., (A), has atoms at the apexes of a cube of side a . Cell of Face Centered, F.C.C., (B), has atoms in
the apexes of cube (as S.C.) plus six atoms in the centers of the cube's faces. Cell of body centered
cube, B.C.C., (C), has atoms arranged in apexes of a cube of side a - as S. C. plus an atom at the
intersection of volume diagonals.
Most metals crystallize into one of three cubic structures described above. Inside a crystal can be
considered different crystal planes. These are parallel planes, identical filled with atoms (or molecules),
separated by equal distances – the interplanar spacing (interplanar distance).
Interplanar spacing is a characteristic of crystalline planes family. Interplanar distances for the cubic
networks have expression d hkl  a h 2  k 2  l 2 ; in expression h, k , l are three integers - indices of
crystalline planes family. When it is irradiated with radiation having a wavelength comparable to the
interplanar distance, the regular structure of crystalline planes, produces diffraction directions separated
from the incident radiation direction by angles 2 as shown in Figure 1. Due to geometric properties,
certain combinations of plane indices lead to the cancellation of the amplitude of diffracted radiation;
accordingly, the respective planes family do not give maximum of diffraction. Thus, in the diffraction
image of a substance which crystallized in the structure B.C.C. exists diffraction maximum only of
families of planes whose indices sum is an even number: 0,1,1 , 2,0,0  , 2,1,1 , 2,2,0  , etc. In the
diffraction image of a substance which crystallized in the structure F.C.C. exists diffraction maximum
only of families of planes whose indices have the same parity (all odd or all even: 1,1,1 , 2,0,0  , 2,2,0  ,
3,1,1 ,
etc. In the diffraction image of a substance which crystallized in the structure S.C. it exists
diffraction maximum for all families of planes whatever indices they have: 1,0,0  , 1,1,0 , 1,1,1 , 2,0,0 ,
etc.
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Above, the sets of crystalline indices were given in order of decreasing of corresponding interplanar
distances.
Figure 1: The formation of the electron diffraction image of the crystal plane of indices hkl 
Using electron microscope only as generator of an electron beam, can be done the diffraction of
electrons on thin conductive polycrystalline layer. Scheme of diffraction experiment is shown in Figure 1.
Monoenergetic electron beam represented by the vertical line, falls on a thin layer composed of
numerous small crystalline grain oriented chaotic. If the beam falls on one of the crystals at the
appropriate angle for diffraction on crystalline plane having indices h, k, l  (represented in the figure by
short, thickened, oblique), then on the direction that makes angle 2 with the direction of the incident
radiation a diffracted beam appears. Diffraction directions due to planes family h, k, l  from all the small
crystals determines a cone of 2 angle. In the experiment, at the distance L from the sample studied
and perpendicular to the initial direction of the electron beam is arranged a photographic plate. The cone
of directions of diffraction causes the apparition on the photographic plate of a circle of radius Rhkl .
Obvious, tg 2   Rhkl L  Dhkl 2L  . In expression, Dhkl is the diameter of the circle appeared on the
photographic plate due to diffraction determined by the planes family of indices h, k, l  from small
crystals in the sample.
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Figure 2: Electron diffraction image of a polycrystalline thin gold foil. Diffraction circles are numbered.
2.2. Requirements for task B
For each of the requirements in this task, write all the answers in the appropriate boxes in the Answer
Sheet.
B.1. Calculate the de Broglie wavelength e of the electrons in the beam diffracted; acceleration voltage
of electrons is U  100 KV . You can use the following values of physical constants: the rest mass of the
electron me  9,11  10 31 kg , the electric charge of the electron e  1,60  10 19 C , the speed of light in
vacuum c  3,00  108 m  s 1 , the Planck constant h  6,63  1034 J  s .
B.2. Determine the difference between the de Broglie wavelength of electrons calculated relativistic and
the de Broglie wavelength of electrons calculated non relativistic.
B.3. Demonstrates that for attached diffraction image, between the diameters Dhkl of diffraction circles
and interplanar spacing d hkl of the crystal planes families responsible for the appearance of these
maxims it is the relationship
Dhkl  d hkl  C  e  L
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and specify the value of constant C .
B.4. Write indices of crystalline planes responsible for the appearance of the first eight diffraction circles
(in ascending order of their diameters so interplanar distances in descending order) for structures S.C.,
F.C.C., B.C.C.. Assume that the side length of the unit cell is a . For each structure calculates the ratio of
the length maximum interplanar distance and length of other interplanar distance and write the results
in appropriate table.
Diffraction image of a very thin film of polycrystalline gold that was made available for you, is enlarged in
aria for 5.3 times. For the experiment conducted L  75 cm .
B.5. Using paper sheet calibrated to task A, measure diameters of diffraction circles; put the results in
appropriate table. Write in the appropriate column of the table in the Answer Sheet the ratio of length of
measured diameter of circles and measured diameter of the smallest circle.
B.6. Analyzes the diameter sizes and relation (1), together with results achieved in the task B.4. and
decides whether the crystal structure of gold is S.C., F.C.C., or B.C.C..
B.7. Plot the dependency between inverses of interplanar distances of planes producing crystalline
diffraction circles (measured in units equal to constant a of network) - and actual diameters of the
corresponding diffraction circles in the image made available. Keep in mind that the image you received
is enlarged.
B.8. Use a graphical analysis of the experimental data set represented in the task B.7. properly using
equation (1) and determines the constant network of gold aAu .
3. Task C
This task proposes you to determine properties of droplets of solution of stearic acid in organic solvent.
Spread onto the surface of water in a watch glass, these droplets form a monomolecular layer. Using a
given data set, you will determine the properties of monolayers.
For each of the requirements in this task , write all the answers in the appropriate boxes in the Answer
Sheet.
We recommend you to carefully read the paragraph 3.1.
3.1. About the formation of a monomolecular layer.
Stearic acid is a saturated fatty acid consisting of a chain of 18 carbon atoms. It is a “waxy solid” with the
chemical formula C17H35CO2H. The Lewis structure of stearic acid (a diagram showing the bonds
between the molecule’s atoms) is presented below. Lines indicate covalent bonds. Each line represents
a pair of electrons which the two atoms are “sharing”.
The acidic character of the substance, and thus the name
of “acid”, is due to the presence of the carboxylic group
COOH at the left end of its chain. This part of the
molecule, which we can informally call its “head”, is electrically
polarized, and can therefore attach to water molecules through
hydrogen bonds. Thus, it is called hydrophilic. The long chain of carbon
and hydrogen atoms that form the right part of the molecule, which we
can informally call the “tail” of the molecule, is the fatter part of the
molecule, which is non-polarizable and thus hydrophobic (repelling
water).
If a stearic acid molecule is placed on the water surface, its polarized
head will attach to the water molecules, while the rest of the stearic acid
molecule – its non-polarizable “tail” – will not be attracted by the water molecules, but rather by the
neighboring stearic acid molecules. Therefore, these interactions cause the stearic acid molecules to
align on the water surface, with their “heads” on the water’s surface and their non-polarized “tails” facing
upwards, away from the water surface. The picture on the left illustrates the combined effect of the
forces acting between the hydrophobic “tails” of the molecule and the forces acting between the
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hydrophilic “heads” and the water molecules. These two effects produce a preferential configuration of
the stearic acid molecules on the surface of water, forming a monomolecular layer.
The picture presents a 3-D representation of stearic acid molecules. The small circles represent
individual atoms (black circles represent carbon atoms, white circles represent hydrogen atoms, and red
circles represent oxygen atoms), while the lines represent bonds between the atoms. As presented in
the picture, we will consider that a stearic acid molecule occupies a L    parallelepiped. If a few
stearic acid molecules are placed on the water surface, they will form a layer with height equal to the
height of one individual molecule. If, instead, many stearic acid molecules are placed on the water
surface, they could pile up on top of each other, yet the first molecules will always form a monomolecular
layer.
The elevation of a stearic acid molecule with respect to the water surface is equal to the length or the
molecule. This length is smaller than the wavelengths of the visible spectrum. Therefore, light shone
upon the layer would not interfere with the molecule, making the monomolecular layer non-visible to the
human eye. However, by adding enough stearic acid molecules, and allowing them to pile up, producing
a thicker layer, the thickest part of the layer can become visible. Alternatively, the layer can also be
observed in the presence of floating talcum particles. The expanding stratum of stearic acid clears the
talcum particles, leaving a circular area free of floating talcum particles.
You may assume the following quantities regarding the stearic acid are known: its molar mass is
 s  284 kg / kmol , density   941kg / m3 , the molecule’s sectional surface is  2  21  10 20 m 2 . The
concentration of the stearic acid solution in the volatile solvent is c  0,02 kg / m 3 . The solvent’s density
is  solvent  878 kg  m 3
3.2. Requirements for task C
C.1. Measurements on droplets
From the set of tools found on the workbench you should use: paper sheet with arbitrary units, marker,
pipette and two test tubes and their support. Test tubes have not a flat base. One of the test tubes
contains the solution of stearic acid in an organic solvent, and the other is empty.
C.1.1. Describe, briefly, a method by which you may determine the volume of a drop delivered by
pipette.
C.1.2. Using the objects available, determines (in m3 ) the volume V p of a drop of solution.
C.1.3. Using paper sheet calibrated in task A, measure the inner diameter d pipeta of the pipette tip. Using
results from C.1.2. estimates value of the surface tension coefficient  of the solution.
C.1.4. Determine (in m3 ) Vs volume of stearic acid in a solution drop, whose volume V p you have
determined in task C.1.2.
C.2. Measurements on monomolecular layers
If water is placed in a watch glass and few drops of
solution studied in C.1 are spread on the surface of
water, a monomolecular layer of stearic acid remains
(after the solvent evaporation) on the surface of the
water - whose diameter Ds can be measured. In the
neighboring picture can be seen the described situation.
The table below presents data on the diameter Ds of
the layer appeared when on water surface is dripping a number N s of drops of stearic acid.
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N s (#)
5
7
9
11
13
15
17
Ds (cm)
4,9
5,7
6,5
7,2
7,9
8,5
9
C.2.1. Properly analyzing the data in the table above determine the expression of height of the layer of
stearic acid and its numerical value.
C.2.2. Considering that 18 carbon atoms of stearic acid molecule are equidistant and disposed vertically
above the water surface, determine the expression of the height of the space occupied by a carbon atom
and its numerical value.
C.3. Estimation of Avogadro's number
C.3.1. Determine the number N m of molecules in each of the layers described in C.2..
C.3.2. Estimate numerical value of Avogadro’s number.
© Problem written by:
Prof. dr. Delia DAVIDESCU
Prof. dr. Ioana STOICA
Prof. dr. Dan Ovidiu CROCNAN
Conf. univ. dr. Adrian DAFINEI
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Answer Sheet
Experimental Problem (20 points)
Polycrystals and monomolecular layers
1. Task A
A.1 Brief description of the experimental method proposed
1.00p
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A.2. Results of measurements made - tables with data collected from measurements
1.50p
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A.3. Processing method of collected experimental data
Determining the value in millimeters of arbitrary unit of length and writing the result in # # mm format.
1.50p
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2. Task B
B.1. Calculation of de Broglie wavelength e of the electrons in the diffracted beam
1.00p
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B.2. Determining the difference between de Broglie wavelength of electrons calculated relativistic and de
Broglie wavelength of electrons calculated non relativistic.
0.50p
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B.3. Demonstration of the fact that for attached diffraction image, between the diameters Dhkl of
diffraction circles and interplanar spacing dhkl of the crystal planes families responsible for the
appearance of these maxima it is the relationship Dhkl  d hkl  C  e  L .Specification of the value of
constant C .
1.00p
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Romanian Master of Physics 2017
B.4. Fulfilling of tables of indices of crystalline planes responsible for the appearance of the first eight
diffraction circles for structures S.C., F.C.C., B.C.C..
S. C. Structure.
Table 1: Description of the crystalline planes of the structure S.C. that give diffraction
maxima
Nr.
h, k , l 
h2  k 2  l 2
h2  k 2  l 2
d hkl
a d hkl
d100 d hkl
1
2
3
1.00p
4
5
6
7
8
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Romanian Master of Physics 2017
B.4. - continuation
B.C.C. structure
Table 2: Description of the crystalline planes of the structure B.C.C. that give diffraction
maxima
Nr.
h, k , l 
h2  k 2  l 2
h2  k 2  l 2
d hkl
a d hkl
d110 d hkl
1
2
1.00p
3
4
5
6
7
8
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B.4. - continuation
F.C.C. Structure
Table 3: Description of the crystalline planes of the structure F.C.C. that give diffraction
maxima
Nr.
h, k , l 
h2  k 2  l 2
h2  k 2  l 2
d hkl
a d hkl
d111 d hkl
1
2
1.00p
3
4
5
6
7
8
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B.5. Measuring and fulfilling of table 4 with data about diameters of diffraction circles and about ratios of
length of measured diameter of circles and measured diameter of the smallest circle
Table 4: Description of diffraction image provided in problem statement
Circle
number
Diameter
u.a.
Di
D1
Diameter
cm
1.00p
B.6. Analyze of results and decision whether the crystal structure of gold is S.C., F.C.C., or B.C.C..C
Table 5
d100 d hkl
d110 d hkl
d111 d hkl
CS
CVC
CFC
Di
D1
1.00p
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B.7. Plot of dependency between inverses of interplanar distances of planes producing crystalline
diffraction circles (measured in units equal to constant a of network) - and actual diameters of the
corresponding diffraction circles in the image made available.
Table 6
Nr.
a d hkl 
h2  k 2  l 2
Dhkl cm
Dhkl cm
(enlarged
image)
(real image)
1
2
3
04
5
6
7
8
1.00p
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B.8. Results of usage of graphical analysis of the experimental data set represented in the task B.7. properly using equation (1) and determination of the network constant of gold aAu .
1.00p
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3. Task C
C.1. Measurements on droplets
C.1.1. Brief description of method used to determine the volume of a drop delivered by pipette.
1.00p
C.1.2. Determination (in m3 ) of volume V p of a drop of solution
1.00p
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C.1.3. Estimation of value of the surface tension coefficient  of the solution
1.00p
C.1.4. Determination (in m3 ) of Vs , volume of stearic acid in a solution drop, whose volume V p you have
determined in task C.1.2.
1,00p
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C.2. Measurements on monomolecular layers
C.2.1. Determination of expression of height of the layer of stearic acid and its numerical value.
1.00p
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C.2.2. Determination of expression and of numerical value of H c , the height of the space occupied by a
carbon atom.
0.50p
C.3. Estimation of Avogadro's number
C.3.1. Determination of number N m of molecules in each of the layers described in C.2..
0.50p
C.3.2. Estimation of numerical value of Avogadro’s number.
0.50p
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